Contracting the stress energy tensor

[dsaun777]

I am trying to understand the scalar form of the Einstein field equations. I know that you can contract the stress-energy tensor using the metric. And for a perfect fluid model, this turns out to be the energy density summed with the pressure. This also gives the Ricci scalar. However, you can also contract with 1-forms. I am a little unsure what the contraction of the tensor with 1-forms means.
I have been working I haven't picked up a physics book in months I am trying to get back into General relativity.

 

[etotheipi]

What is your question? Yes, $T = {T^{\mu}}_{\mu} = T^{\mu \nu} g_{\mu \nu} = (\epsilon + p)u^{\mu} u_{\mu} + p {g^{\mu}}_{\mu} = 3p - \epsilon$ because you can simply evaluate in the local frame where $u^a = (1,0,0,0)$. And then you may take the trace of $G_{ab} = 8 \pi T_{ab}$ to write ${R^{\mu}}_{\mu} - \frac{1}{2} R {g^{\mu}}_{\mu} = 8 \pi {T^{\mu}}_{\mu}$ but again since ${g^{\mu}}_{\mu} = 4$ you just get $R = - 8\pi T$ for the Ricci scalar.

Are you asking more generally about contractions and contracted products? If so, please could you elaborate?

 

[dsaun777]

I'm asking about contractions involving one forms and how they differ from contractions with the metric. They both produce invariant scalars but what is the difference? These are invariants meaning what is density in one frame could take the form of pressure in another frame correct?

 

[etotheipi]

I'm asking about contractions involving one forms and how they differ from contractions with the metric.

It's a general concept, defined on tensors of any rank. Say you have two tensors $P = P^{ab} e_a \otimes e_b$ and $Q = Q_a e^a$. (Here $Q$ is a one-form). First build their tensor product, which in this case will be a rank $(2,1)$ tensor denoted by $R$,$$R:= P \otimes Q = P^{ab} Q_{c} e_a \otimes e_b \otimes e^c := {R^{ab}}_{c} e_a \otimes e_b \otimes e^c$$The contraction $CR$ of $R$ over the two right-most indices is in this case a $(1,0)$ tensor defined by$$(CR)(\omega) = R(\omega, e^{(\mu)}, e_{(\mu)})$$for any dual vector $\omega$. That's really all there is to a contracted product; building the tensor product and contracting the result over a specified pair of indices

They both produce invariant scalars but what is the difference?

You only get a scalar if you contract out all of the indices. Contracting the stress-energy tensor with a one-form still leaves you with an upper index, i.e. a vector.

These are invariants meaning what is density in one frame could take the form of pressure in another frame correct?

No, the proper energy density $\epsilon = T(u_{\mathscr{C}}, u_{\mathscr{C}})$ and pressure $p$ are two scalar fields defined on the spacetime manifold, corresponding to the quantities as measured in the local rest frame of the fluid.

 

[PeterDonis]

I know that you can contract the stress-energy tensor using the metric. And for a perfect fluid model, this turns out to be the energy density summed with the pressure.

No, it doesn't. As @etotheipi showed in post #2, it is the difference of energy density and 3 times the pressure. (Which one has a positive sign and which has a negative sign depends on which metric signature convention you choose; in post #2 the $- + + +$ convention is used. But it is always a difference, not a sum.)

This also gives the Ricci scalar.

Sort of. If you contract the EFE with the metric (which is the same as taking the trace), and assume a perfect fluid stress-energy tensor, you get

$$
- R = 8 \pi T = 8 \pi \left( - \epsilon + 3 p \right)
$$

 

[dsaun777]

No, it doesn't. As @etotheipi showed in post #2, it is the difference of energy density and 3 times the pressure. (Which one has a positive sign and which has a negative sign depends on which metric signature convention you choose; in post #2 the $- + + +$ convention is used. But it is always a difference, not a sum.)Sort of. If you contract the EFE with the metric (which is the same as taking the trace), and assume a perfect fluid stress-energy tensor, you get

$$
- R = 8 \pi T = 8 \pi \left( - \epsilon + 3 p \right)
$$

Well, I meant that it's a sum of pressure terms and a negative energy density.

 

[PeterDonis]

I meant that it's a sum of pressure terms and a negative energy density.

But the energy density isn't negative, it's positive. The minus sign is in the combination of the terms, not in the energy density. Saying "negative energy density" implies that the minus sign is part of the energy density itself.

This illustrates why it's better to use math than vague ordinary language. The math is unambiguous.

 

[pervect]

I am trying to understand the scalar form of the Einstein field equations. I know that you can contract the stress-energy tensor using the metric. And for a perfect fluid model, this turns out to be the energy density summed with the pressure. This also gives the Ricci scalar. However, you can also contract with 1-forms. I am a little unsure what the contraction of the tensor with 1-forms means.
I have been working I haven't picked up a physics book in months I am trying to get back into General relativity.

We had a thread a long while ago where you thought that the often seen expression $\rho + 3P$, for instance from Baez's "The Meaning of Einstein Equation", was equivalent to the trace of the stress energy tensor. But it's not, as we pointed out in that ancient thread, I think that you may still be hanging on to this idea.

Specifically when you say that

I know that you can contract the stress-energy tensor using the metric. And for a perfect fluid model, this turns out to be the energy density summed with the pressure.

you "know" something that isn't true, there is a sign error in this statement. Assuming that it's true leads you into some conceptual errors, I suspect.

I'd suggest forgeting about this entirely for the moment, and reviewing rank 1 tensors. From an index point of view, raising and lowering an index of a rank m tensor consists of taking the tensor product of the rank m tensor with the rank 2 metric tensor, forming a tensor of rank m+2, then using the contraction operation that reduces the rank of a tensor by 2.

You'll need a textbook or source of some sort to study GR and tensors. What text are you using? There are various approaches to the subject, if I use one source to talk about such basic issues, and you are using another, there is very likely to be total confusion and a lack of communication :(.

You might also consider disentangle learning tensors from learning GR. This can be done by considering the tensor approach to electromagnetism, and comparing it to the classical non-tensor approach. Jackson's "Classical Electrodynamics" used to be such a route, but perhaps there are simpler textbooks that take this route nowadays. I rather hope so, Jackson was not an "easy read".

Gaining some physical insight into tensor operations such as raising and lowering index, taking tensor products, and contracting via this route would prepare you to understand the tensor "language" that GR is written in.

 

[dsaun777]

I appreciate the responses but maybe I'm not being clear. Just as you use the diffential distances in the infinitesimal distance equation ds involving the metric you can apply the differentials to contract the SET to get a scalar. How is contracting the SET using the differentials different from contracting using the metric.

 

[etotheipi]

What do you mean by 'contracting the SET using the differentials'?

A differential, e.g. $(\mathrm{d}x^{\mu})_a$, is in fact a one-form (with $\mu$ being a label) and thus it's not mathematically unreasonable to write down something like $s = (\mathrm{d}x^{\mu})_a (\mathrm{d}x^{\nu})_b T^{ab}$ where $s$ is indeed a scalar, but I don't know how you would interpret that; it doesn't seem particularly useful...

 

[PeterDonis]

I don't know how you would interpret that

It's the $\mu$, $\nu$ component of the SET in the chosen coordinate chart. In general this does not have any meaning, but in some cases, if the chart is well chosen, it can be useful.

 

[PeterDonis]

How is contracting the SET using the differentials different from contracting using the metric.

The differentials will change if you change coordinate charts, so the contractions with the differentials are coordinate-dependent. But the contraction with the metric is an invariant; it is the same no matter what coordinate chart you choose.

 

[etotheipi]

It's the $\mu$, $\nu$ component of the SET in the chosen coordinate chart. In general this does not have any meaning, but in some cases, if the chart is well chosen, it can be useful.

Ah, so it is! I had a dumb moment.

 

[dsaun777]

What do you mean by 'contracting the SET using the differentials'?

A differential, e.g. $(\mathrm{d}x^{\mu})_a$, is in fact a one-form (with $\mu$ being a label) and thus it's not mathematically unreasonable to write down something like $s = (\mathrm{d}x^{\mu})_a (\mathrm{d}x^{\nu})_b T^{ab}$ where $s$ is indeed a scalar, but I don't know how you would interpret that; it doesn't seem particularly useful...

This is exactly what I am referring to...

 

[dsaun777]

The differentials will change if you change coordinate charts, so the contractions with the differentials are coordinate-dependent. But the contraction with the metric is an invariant; it is the same no matter what coordinate chart you choose.

So the contraction with the differential 1 forms are not invariant objects necessarily?

 

[etotheipi]

So the contraction with the differential 1 forms are not invariant objects necessarily?

You really need to be more specific. The scalar $T^{\mu \nu} = T^{ab} (\mathrm{d}x^{\mu})_a (\mathrm{d}x^{\nu})_b$ is invariant, but by this we mean it's the $\mu \nu$ component of $T^{ab}$ in a specified co-ordinate chart $x^{\mu}$.

If you change to a different set of co-ordinates $x'^{\mu}$, then $T'^{\mu \nu} = T^{ab} (\mathrm{d}x'^{\mu})_a (\mathrm{d}x'^{\nu})_b$ is the $\mu \nu$ component of $T^{ab}$ in this new chart. And $T^{\mu \nu} \neq T'^{\mu \nu}$ in general. That's the sense in which @PeterDonis means that contractions with differentials are co-ordinate dependent.

But more generally, you're being too vague. A contracted product of any tensors is always an invariant object, precisely because both tensor products and contractions are co-ordinate independent operations. See post #4. If you contract over all of the remaining free indices you get a scalar, whilst if you leave some un-contracted you'll get a higher order tensor (which, as a geometric object is independent of co-ordinates, but has co-ordinate dependent components).

Contracted products with arbitrary one-forms are no different; depending on what you contract it with you might get a scalar, vector, etc., which are all invariant objects geometrically (although, again, remember that tensors of rank $>0$ have co-ordinate dependent components). If $\omega_a$ and $\eta_a$ are any two arbitrary one-forms, then $T^{ab} \omega_a \eta_b$ is a scalar, and invariant.

 

[PeterDonis]

So the contraction with the differential 1 forms are not invariant objects necessarily?

No, they are invariants, just in general not meaningful ones. The contraction with the metric is a meaningful invariant because the metric tensor is the metric tensor; which tensor it is does not depend on your choice of coordinates. But which differential 1-forms you contract with does depend on your choice of coordinates.

 

[Ibix]

So the contraction with the differential 1 forms are not invariant objects necessarily?

A slightly simpler example is the contraction of a vector with a differential one-form: $(dx^\mu)_aV^a$. This would be something like "the projection of $V$ on to my frame's x axis" (or y-axis or whatever). But hopefully here you can see the point @PeterDonis is making: as long as I've communicated which direction I'm calling "my x axis", anyone can determine what I mean by $(dx^\mu)$ and anyone can write it down in any coordinate system they like and compute its contraction with $V$. Everyone will get the same result, but it does depend on my original choice of x axis.

This is a useful thing because if the basis one forms are orthonormal at an observer's location (i.e., are basis one forms for a local Lorentz frame) then their contractions with tensors are things that observer would find useful. For example if $dt_\mu$ is their timelike basis then $dt_\mu dt_\nu T^{\mu\nu}$ is the local energy (or mass) density. As Peter noted, expressed in the observer's coordinates, this turns out to be equal to $T^{tt}$. Sloppy sources (e.g. Wikipedia the last time I looked) will therefore tell you that "the tt component of the stress energy tensor is energy density". Careful sources (e.g. MTW) will specify that they are working in a (possibly only locally) orthonormal basis and tell you that "the contraction of the stress energy tensor twice with the t basis one form is the energy density", because $dt_adt_bT^{ab}$ works for everybody as long as they know which direction t is.

 

[dsaun777]

A slightly simpler example is the contraction of a vector with a differential one-form: $(dx^\mu)_aV^a$. This would be something like "the projection of $V$ on to my frame's x axis" (or y-axis or whatever). But hopefully here you can see the point @PeterDonis is making: as long as I've communicated which direction I'm calling "my x axis", anyone can determine what I mean by $(dx^\mu)$ and anyone can write it down in any coordinate system they like and compute its contraction with $V$. Everyone will get the same result, but it does depend on my original choice of x axis.

This is a useful thing because if the basis one forms are orthonormal at an observer's location (i.e., are basis one forms for a local Lorentz frame) then their contractions with tensors are things that observer would find useful. For example if $dt_\mu$ is their timelike basis then $dt_\mu dt_\nu T^{\mu\nu}$ is the local energy (or mass) density. As Peter noted, expressed in the observer's coordinates, this turns out to be equal to $T^{tt}$. Sloppy sources (e.g. Wikipedia the last time I looked) will therefore tell you that "the tt component of the stress energy tensor is energy density". Careful sources (e.g. MTW) will specify that they are working in a (possibly only locally) orthonormal basis and tell you that "the contraction of the stress energy tensor twice with the t basis one form is the energy density", because $dt_adt_bT^{ab}$ works for everybody as long as they know which direction t is.

How do the units work out? wouldn't that contraction multiply the tensor by units of time squared?

 

[PeterDonis]

How do the units work out? wouldn't that contraction multiply the tensor by units of time squared?

The way @Ibix wrote it in post #18 is somewhat confusing.

The energy density seen by an observer with 4-velocity $u^\mu$ is $T_{\mu \nu} u^\mu u^\nu$. @Ibix was describing this in a coordinate chart in which $u^\mu$ is the timelike basis vector. He called it $dt_\mu$ but that was really a misnomer; it doesn't have units of time. A 4-velocity has no units. So the contraction of it with the stress-energy tensor has the same units as the stress-energy tensor does.

 

[dsaun777]

The way @Ibix wrote it in post #18 is somewhat confusing.

The energy density seen by an observer with 4-velocity $u^\mu$ is $T_{\mu \nu} u^\mu u^\nu$. @Ibix was describing this in a coordinate chart in which $u^\mu$ is the timelike basis vector. He called it $dt_\mu$ but that was really a misnomer; it doesn't have units of time. A 4-velocity has no units. So the contraction of it with the stress-energy tensor has the same units as the stress-energy tensor does.

That makes perfect sense thanks.

 

[dsaun777]

The way @Ibix wrote it in post #18 is somewhat confusing.

The energy density seen by an observer with 4-velocity $u^\mu$ is $T_{\mu \nu} u^\mu u^\nu$. @Ibix was describing this in a coordinate chart in which $u^\mu$ is the timelike basis vector. He called it $dt_\mu$ but that was really a misnomer; it doesn't have units of time. A 4-velocity has no units. So the contraction of it with the stress-energy tensor has the same units as the stress-energy tensor does.

But is that still a differential 1 form or a four-velocity? The four-velocity have no units because it's a tangent vector along a proper length right?

 

[robphy]

The 4-velocity is a “unit”-4-vector… the [timelike] 4-momentum divided by its magnitude…. Hence it is unitless.

 

[dsaun777]

The 4-velocity is a “unit”-4-vector… the [timelike] 4-momentum divided by its magnitude…. Hence it is unitless.

Just direction...

 

[dsaun777]

The 4-velocity is a “unit”-4-vector… the [timelike] 4-momentum divided by its magnitude…. Hence it is unitless.

It doesn't have units of length over time?

 

[robphy]

It doesn't have units of length over time?

In the way most modern relativity texts define it, it is (in spite of its name) dimensionless… it is a unit-vector

(In ordinary geometry, the unit vector $\hat x$ is dimensionless.)

 

[dsaun777]

In the way most modern relativity texts define it, it is (in spite of its name) dimensionless… it is a unit-vector

(In ordinary geometry, the unit vector $\hat x$ is dimensionless.)

But the norm of the four velocity is the speed of light squared.... That has units.

 

[PeterDonis]

is that still a differential 1 form or a four-velocity?

The way I wrote it, it's a 4-velocity vector (one upper index) contracted (twice) with a second-rank covariant tensor (two lower indexes).

The way @Ibix wrote it, it's a 1-form (one lower index) contracted twice with a second-rank contravariant tensor (two upper indexes).

The two ways are equivalent; you can always use the metric to lower the index of the 4-velocity vector to get a 1-form and raise the indexes of the covariant stress-energy tensor to get a contravariant tensor.

Note that the 1-form is not the same thing as a coordinate differential. That's another reason why the way @Ibix wrote it is confusing; despite the notation he used, $dt_\mu$ is not a differential, it's a 1-form. They're two different things. A 1-form is a first rank covariant tensor, which in this case is unitless (because it's obtained by lowering the index of a unitless 4-vector with the metric, which is also unitless). A coordinate differential represents an infinitesimal increment of a coordinate, and has the same units as the coordinate does.

 

[PeterDonis]

the norm of the four velocity is the speed of light squared....

Actually that's the squared norm.

That has units.

Not in the natural units used in relativity. In those units $c = 1$ and is dimensionless, and so are all speeds (speeds in these units represent the ordinary speed divided by $c$).

You could, if you insisted, treat the 4-velocity as having units of speed and construct a consistent system of units that way. But in such a system of units, the stress-energy tensor would have units of energy density divided by $c$ squared, or mass density. (In "natural" units, energy and mass have the same units so this distinction doesn't arise.)

 

[vanhees71]

It doesn't have units of length over time?

It's very convenient to normalize it to 1. The four-velocity of a massive particle is thus
$$u^{\mu}=\frac{1}{c} \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau},$$
which by definition of proper time is normalized to 1,
$$g_{\mu \nu} u^{\mu} u^{\nu}=1,$$
when using the west-coast convention for the pseudo-metric, i.e., the signature (1,-1,-1,-1).

Another even more convenient choice is to use natural units with $\hbar=c=k_{\text{B}}=1$ (and maybe even $G=1$, which then makes everything measured in dimensionless quantities, i.e., Planck units).

 

[dsaun777]

It's very convenient to normalize it to 1. The four-velocity of a massive particle is thus
$$u^{\mu}=\frac{1}{c} \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau},$$
which by definition of proper time is normalized to 1,
$$g_{\mu \nu} u^{\mu} u^{\nu}=1,$$
when using the west-coast convention for the pseudo-metric, i.e., the signature (1,-1,-1,-1).

Another even more convenient choice is to use natural units with $\hbar=c=k_{\text{B}}=1$ (and maybe even $G=1$, which then makes everything measured in dimensionless quantities, i.e., Planck units).

Yes thanks for the clarification. But there are still units involved even though it's length over length I still consider those units because it relates to physical measurements.

 

[robphy]

Yes thanks for the clarification. But there are still units involved even though it's length over length I still consider those units because it relates to physical measurements.

So, would you say the following?

In Euclidean geometry, the unit-vector along the $x$-direction $\hat x$ is not dimensionless… but has units of m/m … or, if I were describing the direction of a force in classical physics, that unit vector along the force has units N/N or maybe $N/\sqrt{N^2}$.

 

[dsaun777]

So, would you say the following?

In Euclidean geometry, the unit-vector along the $x$-direction $\hat x$ is not dimensionless… but has units of m/m … or, if I were describing the direction of a force in classical physics, that unit vector along the force has units N/N or maybe $N/\sqrt{N^2}$.

If you are doing calculations on paper in an abstract manner to teach a class, I would say that you are not using any units. But if you are using applied physics to solve a real engineering problem it would be useful, I think, to include units even if they cancel and become dimensionless. If anything to show where unitless quantities get their respective derivation and meaning.

 

[Maarten Havinga]

I like the way PF members are discussing physics here, it goes somewhere and I feel space for misunderstandings to be solved. Thanks for that everyone! :)